Question: You have found the following ages (in years) of all 6 lizards at your local zoo: $ 2,\enspace 1,\enspace 2,\enspace 1,\enspace 1,\enspace 2$ What is the average age of the lizards at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 1 + 2 + 1 + 1 + 2}{{6}} = {1.5\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $0.5$ years $0.25$ years $^2$ $1$ year $-0.5$ years $0.25$ years $^2$ $2$ years $0.5$ years $0.25$ years $^2$ $1$ year $-0.5$ years $0.25$ years $^2$ $1$ year $-0.5$ years $0.25$ years $^2$ $2$ years $0.5$ years $0.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.25} + {0.25} + {0.25} + {0.25} + {0.25} + {0.25}} {{6}} $ $ {\sigma^2} = \dfrac{{1.5}}{{6}} = {0.25\text{ years}^2} $ The average lizard at the zoo is 1.5 years old. The population variance is 0.25 years $^2$.